Question

if the sum of first p terms of an ap is a p square + b p find its common difference​

Answer 1

Answer:

Given,

$\sf{S_p} = \sf{ap^2} + \sf{bp}$

We know,

$\sf{S_n} = \sf \frac{n}{2} [2a+(n-1)d]$

Where d is common difference

As per your question,

$\sf{S_p} = \sf \frac{p}{2} [2a+(p-1)d]$

$\sf{S_p} = \sf{2ap} + \sf \frac{dp}{2} [p-1]$

$\sf{S_p} = \sf{2ap} + \Large{\sf \frac{dp^2}{2}} - \sf{dp}{2}$

From the question we know $\sf{S_n}$ value so

$\sf{ap^2} + \sf{bp} = \sf{2ap} + \sf \frac{dp^2}{2} - \sf{dp}{2}$

$\sf{ap^2} + \sf{bp} = \Large{\sf \frac{dp^2}{2}} + \sf{(2a - \Large{\frac{d}{2}})p}$

From the above equations, we can see

a = $\Large \sf \frac{d}{2}$

b = $\sf{2a} - \Large{ \sf \frac{d}{2}}$

b = $\sf{2a} - \sf{a}$

b = a

And we know,

a = $\Large \sf \frac{d}{2}$

d = $\sf{2a}$